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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p>Next, suppose that we impose the initial conditions</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_5.html ./knowl/eq3_5_2.html">
\begin{equation}
y(x_0)=y_0,\quad y^{\prime}(x_0)=y_1.\tag{3.1.9}
\end{equation}
</div>
<p class="continuation">Then we have</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_5.html ./knowl/eq3_5_2.html">
\begin{equation*}
\begin{aligned}
&amp;y(x_0)=y_0: ~ y_0=C_1 e^{r_1 x_0}+C_2 e^{r_2 x_0};\\
&amp; y^{\prime}(x_0)=y_1: ~ y_1=C_1 r_1 e^{r_1 x_0}+C_2 r_2 e^{r_2 x_0}.\\
&amp;\rightarrow C_1=\frac{y_1-y_0 r_2}{r_1-r_2} e^{-r_1 x_0},\quad C_2=\frac{y_0 r_1-y_1}{r_1-r_2} e^{-r_2 x_0}.
\end{aligned}
\end{equation*}
</div>
<p class="continuation">Since <span class="process-math">\(C_1\)</span> and <span class="process-math">\(C_2\)</span> are uniquely determined, we have the unique solution to the initial value problem (<a href="" class="xref" data-knowl="./knowl/eq3_5.html" title="Equation 3.1.6">(3.1.6)</a>) and (<a href="" class="xref" data-knowl="./knowl/eq3_5_2.html" title="Equation 3.1.9">(3.1.9)</a>), which is</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_5.html ./knowl/eq3_5_2.html">
\begin{equation*}
y=\frac{y_1-y_0 r_2}{r_1-r_2} e^{-r_1 x_0} e^{r_1 x}+\frac{y_0 r_1-y_1}{r_1-r_2} e^{-r_2 x_0} e^{r_2 x}.
\end{equation*}
</div>
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